Limit of a Harmonic Sum

Let $L_n$ be defined as follows:

$L_n = H_n - \frac{1}{2^n} \sum_{r=1}^{n} {n \choose r} H_r$

So, we need to find $\lim_{n \to \infty} L_n$

Now, using the integral representation of $H_k$ and binomial theorem, $\begin{aligned} L_n &= \int_0^1 \frac{1-x^n}{1-x} \, dx - \int_0^1 \frac{1-\left(\frac{1+x}{2} \right)^n }{1-x} \, dx \\&= \int_0^1 \frac{\left(1-\frac{x}{2} \right)^n - (1-x)^n}{x} \, dx \quad (\text{by using } x \mapsto 1-x) \\&= \int_1^2 \frac{\partial}{\partial a} \int_0^1 \frac{\left( 1- \frac{x}{a} \right)^n}{x} \, dx \, da \\&= \int_1^2 \int_0^1 \frac{\partial}{\partial a} \frac{\left( 1- \frac{x}{a} \right)^n}{x} \, dx \, da \\&= \int_1^2 \frac{n}{n+1} \frac{1}{a} \left\{ 1- \left( 1- \frac{1}{a} \right)^{n+1} \right \} \, da\end{aligned}$

Note that the integrand in the last line is dominated by $1$ (in the relevant interval). Using dominated convergence, we have

$\lim_{n \to \infty} L_n = \int_1^2 \frac{da}{a} = \ln \boxed{2}$

Since $H_n \; = \; \sum_{k=1}^n \frac{(-1)^{k-1}}{k}\binom{n}{k}$ we have $\sum_{r=1}^n \binom{n}{r}H_r \; = \; \sum_{k=1}^n \frac{(-1)^{k-1}}{k}\sum_{r=k}^n \binom{n}{r}\binom{r}{k} \; = \; \sum_{k=1}^n \frac{(-1)^{k-1}}{k}2^{n-k}\binom{n}{k}$ and hence $\begin{array}{rcl} \displaystyle L_n \; = \; H_n - 2^{-n}\sum_{r=1}^n \binom{n}{r}H_r & = & \displaystyle \sum_{k=1}^n \frac{(-1)^{k-1}}{k}\big[1 - 2^{-k}\big]\binom{n}{k} \; = \; \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \int_{\frac12}^1 t^{k-1}\,dt \\ & = & \displaystyle \int_{\frac12}^1 \frac{1 - (1-t)^n}{t}\,dt \; = \; \ln2 - \int_{\frac12}^1 (1-t)^n\frac{dt}{t} \end{array}$ and hence $\left|L_n - \ln2\right| \; \le\; \frac{\ln2}{2^n}$ Thus the answer is $2^2 = \boxed{4}$ , but we also see that the rate of convergence is exponential.